Optimal. Leaf size=282 \[ -\frac{1}{2} a e \text{PolyLog}\left (2,-c^2 x^2\right )-\frac{1}{2} i b e \left (-\log \left (c^2 x^2+1\right )+\log (1-i c x)+\log (1+i c x)\right ) \text{PolyLog}(2,-i c x)+\frac{1}{2} i b e \left (-\log \left (c^2 x^2+1\right )+\log (1-i c x)+\log (1+i c x)\right ) \text{PolyLog}(2,i c x)+\frac{1}{2} i b d \text{PolyLog}(2,-i c x)-\frac{1}{2} i b d \text{PolyLog}(2,i c x)-i b e \text{PolyLog}(3,1-i c x)+i b e \text{PolyLog}(3,1+i c x)+i b e \log (1-i c x) \text{PolyLog}(2,1-i c x)-i b e \log (1+i c x) \text{PolyLog}(2,1+i c x)+a d \log (x)+\frac{1}{2} i b e \log (i c x) \log ^2(1-i c x)-\frac{1}{2} i b e \log (-i c x) \log ^2(1+i c x) \]
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Rubi [A] time = 0.341867, antiderivative size = 282, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {5015, 4848, 2391, 5013, 5011, 2396, 2433, 2374, 6589} \[ -\frac{1}{2} a e \text{PolyLog}\left (2,-c^2 x^2\right )-\frac{1}{2} i b e \left (-\log \left (c^2 x^2+1\right )+\log (1-i c x)+\log (1+i c x)\right ) \text{PolyLog}(2,-i c x)+\frac{1}{2} i b e \left (-\log \left (c^2 x^2+1\right )+\log (1-i c x)+\log (1+i c x)\right ) \text{PolyLog}(2,i c x)+\frac{1}{2} i b d \text{PolyLog}(2,-i c x)-\frac{1}{2} i b d \text{PolyLog}(2,i c x)-i b e \text{PolyLog}(3,1-i c x)+i b e \text{PolyLog}(3,1+i c x)+i b e \log (1-i c x) \text{PolyLog}(2,1-i c x)-i b e \log (1+i c x) \text{PolyLog}(2,1+i c x)+a d \log (x)+\frac{1}{2} i b e \log (i c x) \log ^2(1-i c x)-\frac{1}{2} i b e \log (-i c x) \log ^2(1+i c x) \]
Antiderivative was successfully verified.
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Rule 5015
Rule 4848
Rule 2391
Rule 5013
Rule 5011
Rule 2396
Rule 2433
Rule 2374
Rule 6589
Rubi steps
\begin{align*} \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x} \, dx &=d \int \frac{a+b \tan ^{-1}(c x)}{x} \, dx+e \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (1+c^2 x^2\right )}{x} \, dx\\ &=a d \log (x)+\frac{1}{2} (i b d) \int \frac{\log (1-i c x)}{x} \, dx-\frac{1}{2} (i b d) \int \frac{\log (1+i c x)}{x} \, dx+(a e) \int \frac{\log \left (1+c^2 x^2\right )}{x} \, dx+(b e) \int \frac{\tan ^{-1}(c x) \log \left (1+c^2 x^2\right )}{x} \, dx\\ &=a d \log (x)+\frac{1}{2} i b d \text{Li}_2(-i c x)-\frac{1}{2} i b d \text{Li}_2(i c x)-\frac{1}{2} a e \text{Li}_2\left (-c^2 x^2\right )+\frac{1}{2} (i b e) \int \frac{\log ^2(1-i c x)}{x} \, dx-\frac{1}{2} (i b e) \int \frac{\log ^2(1+i c x)}{x} \, dx+\left (b e \left (-\log (1-i c x)-\log (1+i c x)+\log \left (1+c^2 x^2\right )\right )\right ) \int \frac{\tan ^{-1}(c x)}{x} \, dx\\ &=a d \log (x)+\frac{1}{2} i b e \log (i c x) \log ^2(1-i c x)-\frac{1}{2} i b e \log (-i c x) \log ^2(1+i c x)+\frac{1}{2} i b d \text{Li}_2(-i c x)-\frac{1}{2} i b d \text{Li}_2(i c x)-\frac{1}{2} a e \text{Li}_2\left (-c^2 x^2\right )-(b c e) \int \frac{\log (i c x) \log (1-i c x)}{1-i c x} \, dx-(b c e) \int \frac{\log (-i c x) \log (1+i c x)}{1+i c x} \, dx+\frac{1}{2} \left (i b e \left (-\log (1-i c x)-\log (1+i c x)+\log \left (1+c^2 x^2\right )\right )\right ) \int \frac{\log (1-i c x)}{x} \, dx-\frac{1}{2} \left (i b e \left (-\log (1-i c x)-\log (1+i c x)+\log \left (1+c^2 x^2\right )\right )\right ) \int \frac{\log (1+i c x)}{x} \, dx\\ &=a d \log (x)+\frac{1}{2} i b e \log (i c x) \log ^2(1-i c x)-\frac{1}{2} i b e \log (-i c x) \log ^2(1+i c x)+\frac{1}{2} i b d \text{Li}_2(-i c x)-\frac{1}{2} i b e \left (\log (1-i c x)+\log (1+i c x)-\log \left (1+c^2 x^2\right )\right ) \text{Li}_2(-i c x)-\frac{1}{2} i b d \text{Li}_2(i c x)+\frac{1}{2} i b e \left (\log (1-i c x)+\log (1+i c x)-\log \left (1+c^2 x^2\right )\right ) \text{Li}_2(i c x)-\frac{1}{2} a e \text{Li}_2\left (-c^2 x^2\right )+(i b e) \operatorname{Subst}\left (\int \frac{\log (x) \log \left (-i c \left (\frac{i}{c}-\frac{i x}{c}\right )\right )}{x} \, dx,x,1+i c x\right )-(i b e) \operatorname{Subst}\left (\int \frac{\log (x) \log \left (i c \left (-\frac{i}{c}+\frac{i x}{c}\right )\right )}{x} \, dx,x,1-i c x\right )\\ &=a d \log (x)+\frac{1}{2} i b e \log (i c x) \log ^2(1-i c x)-\frac{1}{2} i b e \log (-i c x) \log ^2(1+i c x)+\frac{1}{2} i b d \text{Li}_2(-i c x)-\frac{1}{2} i b e \left (\log (1-i c x)+\log (1+i c x)-\log \left (1+c^2 x^2\right )\right ) \text{Li}_2(-i c x)-\frac{1}{2} i b d \text{Li}_2(i c x)+\frac{1}{2} i b e \left (\log (1-i c x)+\log (1+i c x)-\log \left (1+c^2 x^2\right )\right ) \text{Li}_2(i c x)-\frac{1}{2} a e \text{Li}_2\left (-c^2 x^2\right )+i b e \log (1-i c x) \text{Li}_2(1-i c x)-i b e \log (1+i c x) \text{Li}_2(1+i c x)-(i b e) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,1-i c x\right )+(i b e) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,1+i c x\right )\\ &=a d \log (x)+\frac{1}{2} i b e \log (i c x) \log ^2(1-i c x)-\frac{1}{2} i b e \log (-i c x) \log ^2(1+i c x)+\frac{1}{2} i b d \text{Li}_2(-i c x)-\frac{1}{2} i b e \left (\log (1-i c x)+\log (1+i c x)-\log \left (1+c^2 x^2\right )\right ) \text{Li}_2(-i c x)-\frac{1}{2} i b d \text{Li}_2(i c x)+\frac{1}{2} i b e \left (\log (1-i c x)+\log (1+i c x)-\log \left (1+c^2 x^2\right )\right ) \text{Li}_2(i c x)-\frac{1}{2} a e \text{Li}_2\left (-c^2 x^2\right )+i b e \log (1-i c x) \text{Li}_2(1-i c x)-i b e \log (1+i c x) \text{Li}_2(1+i c x)-i b e \text{Li}_3(1-i c x)+i b e \text{Li}_3(1+i c x)\\ \end{align*}
Mathematica [F] time = 0.204386, size = 0, normalized size = 0. \[ \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x} \, dx \]
Verification is Not applicable to the result.
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Maple [C] time = 2.036, size = 6931, normalized size = 24.6 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} a d \log \left (x\right ) + \frac{1}{2} \, \int \frac{2 \,{\left (b d \arctan \left (c x\right ) +{\left (b e \arctan \left (c x\right ) + a e\right )} \log \left (c^{2} x^{2} + 1\right )\right )}}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b d \arctan \left (c x\right ) + a d +{\left (b e \arctan \left (c x\right ) + a e\right )} \log \left (c^{2} x^{2} + 1\right )}{x}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )}{\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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